Question

The integral $\int sec^{2/3} x cosec^{4/3}x\, dx$ is equal to (Hence $C$ is a constant of integration)

• A. $3 tan^{-1/3}x + C$
• B. $-\frac{3}{4} tan^{-4/3} x +C$
• C. $-3 cot^{-1/3}x +c$
• D. $-3 tan^{-1/3}x +C$

Answer 1

Answer: (D)

$-3 tan^{-1/3}x +C$

Answer 2

The correct answer is D:$I = \frac{-3}{(tan x)^{1/3}}+c$
$I = \int \frac{dx}{(sin x)^{4/3} . (cos x)^{2/3}}$
$I = \int \frac{dx}{\bigg(\frac{sin x}{cos x}\bigg)^{4/3} . cos^2 x}$
$\Rightarrow \, I = \int \frac{sec^2 x}{(tan x)^{4/3}} dx$
put tanx = t $\Rightarrow \, \, sec^2 x dx = dt$
$\therefore \, \, I = \int \frac{dt}{t^{4/3}} \Rightarrow \, I = \frac{-3}{t^{1/3}} +c$
$\Rightarrow \, \, \, I = \frac{-3}{(tan x)^{1/3}}+c$